What is the foci ((X-1)^2)/4+((Y+2)^2)/(16)=1 ?

The foci ((X-1)^2)/4+((Y+2)^2)/(16)=1 is (1,-2+2sqrt(3)),(1,-2-2sqrt(3))

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foci ((X-1)^2)/4+((Y+2)^2)/(16)=1

foci

(1, - 2 + 23), (1, - 2 - 23)

Solution steps

(h, k + c), (h, k - c)

Calculate ellipse properties

\frac{( X - 1 ) ^{2}}{4} + \frac{( Y + 2 ) ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (1, - 2), b = 4, a = 2

(1, - 2 + c), (1, - 2 - c)

Compute

c :

c = 4^{2} - 2^{2} : 23

(1, - 2 + 23), (1, - 2 - 23)

What is the foci ((X-1)^2)/4+((Y+2)^2)/(16)=1 ?
The foci ((X-1)^2)/4+((Y+2)^2)/(16)=1 is (1,-2+2sqrt(3)),(1,-2-2sqrt(3))